Calculating Velocity of Student in a Horizontal Throw | Physics Force Question

  • Thread starter Thread starter moe6142
  • Start date Start date
  • Tags Tags
    Force Physics
AI Thread Summary
To find the velocity of the student when the object was released, conservation of momentum must be applied. The initial momentum of the system is zero, as both the student and the object are at rest before the throw. When the 3.0 kg object is thrown with a final velocity of 9.6 m/s, its momentum is calculated as 3.0 kg multiplied by 9.6 m/s. To maintain the conservation of momentum, the student's momentum must equal the negative of the object's momentum, leading to the equation: (mass of student) x (velocity of student) = - (3.0 kg x 9.6 m/s). Solving this gives the student's velocity, ensuring that momentum remains conserved in the system.
moe6142
Messages
5
Reaction score
0

Homework Statement



while standing on a horizontal frictionless surface, a 45kg student throws a 3.0kg object to her right. during the throw the object was accelerated horizontally through a distance of 0.6meters from rest to a velocity of 9.6m/s.

what was the velocity of the student when the object was released?
 
Physics news on Phys.org
moe6142 said:

Homework Statement



while standing on a horizontal frictionless surface, a 45kg student throws a 3.0kg object to her right. during the throw the object was accelerated horizontally through a distance of 0.6meters from rest to a velocity of 9.6m/s.

what was the velocity of the student when the object was released?

momentum must be conserved.

p = mv
P_i=P_f

Remember that momentum is a vector, so signs matter. This will be used in this solution when you try to find the velocity of the student that would bring the final sum to zero.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top